A meridian disk of the solid torus wraps q times around its image disk. Here p =1 and q =2.

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1 4. HYPERBOLIC DEHN SURGERY A meridian disk of the solid torus wraps q times around its image disk. Here p =1 and q =2. Any three-manifold with a complete codimension-2 hyperbolic foliation has universal cover H 2 R, and covering transformations act as global isometries in the first coordinate. Because of this strong structure, we can give a complete classification of such manifolds. A Seifert fibration of a three-manifold M is a projection p : M B to some surface B, sothatp is a submersion and the preimages of points are circles in M. A Seifert fibration is a fibration except at a certain number of singular points x 1,...,x k. The model for the behavior in p 1 (N ɛ (x i )) is a solid torus with a foliation having the core circle as one leaf, and with all other leaves winding p times around the short way and q times around the long way, where 1 <p<qand (p, q) =1. The projection of a meridian disk of the solid torus to its image in B is q-to-one, 4.28 except at the center where it is one-to-one. A group of isometries of a Riemannian manifold is discrete if for any x, the orbit of x intersects a small neighborhood of x only finitely often. A discrete group Γ of orientation-preserving isometries of a surface N has quotient N/Γ a surface. The projection map N N/Γ is a local homeomorphism except at points x where the isotropy subgroup Γ x is nontrivial. In that case, Γ x is a cyclic group Z/qZ for some q>1, and the projection is similar to the projection of a meridian disk cutting across a singular fiber of a Seifert fibration. Theorem 4.9. Let F be a hyperbolic foliation of a closed three-manifold M. Then either Thurston The Geometry and Topology of 3-Manifolds

2 4.8. DEGENERATION OF HYPERBOLIC STRUCTURES. (a) The holonomy group H(π 1 M) is a discrete group of isometries of H 2, and the developing map goes down to a Seifert fibration D /π1 M : M H 2 /H(π 1 M), or (b) The holonomy group is not discrete, and M fibers over the circle with fiber a torus. The structure of F and M in case (b) will develop in the course of the proof Proof. (a) If H(π 1 M) is discrete, then H 2 /H(π 1 M) is a surface. Since M is compact the fibers of the fibration D : M 3 H 2 are mapped to circles under the projection π : M 3 M 3. It follows that D/H(π 1 M):M 3 H 2 /H(π 1 M)isa Seifert fibration. (b) When H(π 1 M) is not discrete, the proof is more involved. First, let us assume that the foliation is oriented (this means that the leaves of the foliation are oriented, or in other words, it is determined by a vector field). We choose a π 1 M-invariant Riemannian metric g in M 3 and let τ be the plane field which is perpendicular to the fibers of D : M 3 H 2. We also insist that along τ, g be equal to the pullback of the hyperbolic metric on H 2. By construction, g defines a metric on M 3, and, since M 3 is compact, there is an infimum I to the length of a nontrivial simple closed curve in M 3 when measured with respect to g. Giveng 1,g 2 π 1 M, we say that they are comparable if there is a y M 3 such that d ( D(g 1 (y)), D(g 2 (y)) ) <I, where d(, ) denotes the hyperbolic distance in H 2. In this case, take the geodesic in H 2 from D(g 1 (y)) to D(g 2 (y)) and look at its horizontal lift at g 2 (y). Suppose its other endpoint e where g 1 (y). Then the length of the lifted path would be equal to the length of the geodesic in H 2, which is less than I. Sinceg 1 g2 1 takes g 2 (y) to g 1 (y), the path represents a nontrivial element of π 1 M and we have a contradiction. Now if we choose a trivialization of H 2 R, we can decide whether or not g 1 (x) is greater than e. If it is greater than e we say that g 1 is greater than g 2,andwrite 4.30 g 1 >g 2, otherwise we write g 1 <g 2. To see that this local ordering does not depend on our choice of y, we need to note that U(g 1,g 2 )={x d ( H(g 1 (x)), H(g 2 (x)) ) <I} is a connected (in fact convex) set. This follows from the following lemma, the proof of which we defer. Lemma f g1,g 2 (x) =d(g 1 x, g 2 x) is a a convex function on H 2. Thurston The Geometry and Topology of 3-Manifolds 65

3 4. HYPERBOLIC DEHN SURGERY One useful property of the ordering is that it is invariant under left and right multiplication. In other words g 1 <g 2 if and only if, for all g 3,wehaveg 3 g 1 <g 3 g 2 and g 1 g 3 <g 2 g 3. To see that the property of comparability is equivalent for these three pairs, note that since H(π 1 H 3 ) acts as isometries on H 2, d(dg 1 y,dg 2 y) <I implies that d(dg 3 g 1 y, Dg 3 g 2 y) <I. Also, if d(dg 1 y,dg 2 y)thend(dg 3 g 1 (g3 1 y),dg 3 g 2 (g3 1 y)) <I,sothatg 1 g 3 and g 2 g 3 are comparable. The invariance of the ordering easily follows (using the fact that π 1 M preserves orientation of the R factors). For a fixed x H 2 we let G ɛ (X) π 1 M be those elements whose holonomy acts on x in a way C 1 ɛ-close to the identity. In other words, for g G ɛ (x), d(x, H g (x)) <ɛand the derivative of H g (x) parallel translated back to x, isɛ-close to the identity. Proposition There is an ɛ 0 so that for all ɛ<ɛ 0 [G ɛ,g ɛ ] G ɛ. Proof. For any Lie group the map [, ] :G G G has derivative zero at (id, id). Since for any g G, (g, id) id and (id,g) 1. The tangent spaces of G id and id G span the tangent space to G G at (id, id). Apply this to the group of isometries of H 2. From now on we choose ɛ<i/8so that any two words of length four or less in G ɛ are comparable. We claim that there is some β G ɛ which is the smallest element in G ɛ which is > id. In other words, if id <α G ɛ, a β, thenα>β.thiscanbe seen as follows. Take an ɛ-ball B of x H 2 and look at its inverse image B under D. Choose a point y in B and consider y and α(y), where α G ɛ. We can truncate B by the lifts of B (using the horizontal lifts of the geodesics through x) through y and α(y). Since this is a compact set there are only a finite number of images of y under π 1 M contained in it. Hence there is one β(y) whoser coordinate is the closest to that of y itself. β is clearly our minimal element Now consider α>β>1, α G ɛ. By invariance under left and right multiplication, α 2 >β α >αand α>α 1 βα > 1. Suppose α 1 βα < β. Thenβ>α 1 βα > 1 so that 1 >α 1 βαβ 1 >β 1. Similarly if α 1 βα > β > 1thenβ>αβα 1 > 1 so that 1 >αβα 1 β 1 >β 1. Note that by multiplicative invariance, if g 1 >g 2 then g 1 2 = g 1 1 g 1 g 1 2 >g 1 1 g 2 g 1 2 = g 1 1. Wehaveeither1<βα 1 β 1 α<βor 1 <βαβ 1 α 1 <βwhich contradicts the minimality of β.thusα 1 βα = β for all α G ɛ. We need to digress here for a moment to classify the isometries of H 2. We will prove the following: Proposition If g : H 2 H 2 is a non-trivial isometry of H 2 which preserves orientation, then exactly one of the following cases occurs: 66 Thurston The Geometry and Topology of 3-Manifolds

4 4.8. DEGENERATION OF HYPERBOLIC STRUCTURES. There are only finitely many translates of y in this region. (i) g has a unique interior fixed point or (ii) g leaves a unique invariant geodesic or (iii) g has a unique fixed point on the boundary of H 2. Case (i) is called elliptic, case (ii) hyperbolic, case (iii) parabolic. Proof. This follows easily from linear algebra, but we give a geometric proof. Pick an interior point x H 2 and connect x to gx by a geodesic l 0. Draw the geodesics l 1, l 2 at gx and g 2 x which bisect the angle made by l 0 and gl 0, gl 0 and g 2 l 0 respectively. There are three cases: 4.33 (i) l 1 and l 2 intersect in an interior point y (ii) There is a geodesic l 3 perpendicular to l 1, l 2 (iii) l 1,l 2 are parallel, i.e., they intersect at a point at infinity x 3. Thurston The Geometry and Topology of 3-Manifolds 67

5 4. HYPERBOLIC DEHN SURGERY In case (i) the length of the arc gx, y equals that of g 2 x, y since (gx, g 2 x, y) is an isoceles triangle. It follows that y is fixed by g. In case (ii) the distance from gx to l 3 equals that from g 2 x to l 3.Sincel 3 meets l 1 and l 2 in right angles it follows that l 3 is invariant by g Finally, in case (iii) g takes l 1 and l 2, both of which hit the boundary of H 2 in the same point x 3. It follows that g fixes x 3 since an isometry takes the boundary to itself. Uniqueness is not hard to prove. Using the classification of isometries of H 2, it is easy to see that the centralizer of any non-trivial element g in isom(h 2 ) is abelian. (For instance, if g is elliptic with fixed point x 0, then the centralizer of g consists of elliptic elements with fixed point x 0 ). It follows that the centralizer of β in π 1 (M) is abelian; let us call this group N. Although G ɛ (x) depends on the point x, for any point x H 2, if we choose ɛ small enough, then G ɛ (x ) G ɛ (x). In particular if x = H(g)x, g π 1 M,then all elements of G ɛ (x ) commute with β. It follows that N is a normal subgroup of π 1 (M). Consider now the possibility that β is elliptic with fixed point x 0 and n N fixes x 0 we see that all of π 1 M must fix x 0. But the function f x0 : H 2 R + which 68 Thurston The Geometry and Topology of 3-Manifolds

6 4.8. DEGENERATION OF HYPERBOLIC STRUCTURES. measures the distance of a point in H 2 from x 0 is H(π 1 M) invariant so that it lifts toafunctionf on M 3. However, M 3 is compact and the image of f is non-compact, which is impossible. Hence β cannot be elliptic. If β were hyperbolic, the same reasoning would imply that H(π 1 M)leavesinvari ant the invariant geodesic of β. In this case we could define f l : H 2 R to be the distance of a point from l. Again, the function lifts to a function on M 3 and we have a contradiction. The case when β is parabolic actually does occur. Let x 0 be the fixed point of β on the circle at infinity. N must also fix x 0. Using the upper half-plane model for H 2 with x 0 at, β acts as a translation of R 2 and N must act as a group of similarities; but since they commute with β, they are also translations. Since N is normal, π 1 M must act as a group of similarities of R 2 (preserving the upper half-plane). Clearly there is no function on H 2 measuring distance from the point x 0 at infinity. If we consider a family of finite points x τ X, and consider the functions f xτ,even though f xτ blows up, its derivative, the closed 1-form df xτ, converges to a closed 1 form ω. Geometrically, ω vanishes on tangent vectors to horocycles about x 0 and takes the value 1 on unit tangents to geodesics emanating from x 0. The non-singular closed 1-form ω on H 2 is invariant by H (π 1 M), hence it defines a non-singular closed one-form ω on M. The kernel of ω is the tangent space to the leaves of a codimension one foliation F of M. The leaves of the corresponding foliation F on M are the preimages under D of the horocycles centered at x 0. The group of periods for ω must be discrete, for otherwise there would be a translate of the horocycle about x 0 through x close to x, hence an element of G ɛ which does not commute with β. Let p 0 be the smallest period. Then integration of ω defines a map from M to S 1 = R/ p 0, which is a fibration, with fibers the leaves of F. The fundamental group of each fiber is contained in N, which is abelian, so the fibers are toruses Thurston The Geometry and Topology of 3-Manifolds 69

7 4. HYPERBOLIC DEHN SURGERY It remains to analyze the case that the hyperbolic foliation is not oriented. In this case, let M bethedoublecoverofm which orients the foliation. M fibers over S 1 with fibration defined by a closed one-form ω. Sinceω is determined by the unique fixed point at infinity of H(π 1 M ), ω projects to a non-singular closed one-form on M. This determines a fibration of M with torus fibers. (Klein bottles cannot occur even if M is not required to be orientable.) We can construct a three-manifold of type (b) by considering a matrix A SL(2, Z) which is hyperbolic, i.e., it has two eigenvalues λ 1,λ 2 and two eigenvectors V 1,V 2. Then AV 1 = λ 1 V 1,AV 2 = λ 2 V 2 and λ 2 =1/λ 1. Since A SL(2, Z) preserves Z Z its action on the plane descends to an action on the torus T 2. Our three-manifold M A is the mapping torus of the action of A 4.37 on T 2. Notice that the lines parallel to V 1 are preserved by A so they give a onedimensional foliation on M A. Of course, the lines parallel to V 2 also define a foliation. The reader may verify that both these foliations are hyperbolic. When A = [ then M A is the manifold (S 3 K) (D,±1) obtained by Dehn surgery on the figure-eight knot. The hyperbolic foliations corresponding to (0, 1) and (0, 1) are distinct, and they correspond to the two eigenvectors of [ ]. All codimension-2 hyperbolic foliations with leaves which are not closed are obtained by this construction. This follows easily from the observation that the hyperbolic foliation restricted to any fiber is given by a closed non-singular one-form, together with the fact that a closed non-singular one-form on T 2 is determined (up to isotopy) by its cohomology class. The three manifolds (S 3 K) (1,1), (S 3 K) (2,1) and (S 3 K) (3,1) also have codimension-2 hyperbolic foliations which arise as limits of hyperbolic structures. Since they are rational homology spheres, they must be Seifert fiber spaces. A Seifert fiber space cannot be hyperbolic, since (after passing to a cover which orients the fibers) a general fiber is in the center of its fundamental group. On the other hand, the centralizer of an element in the fundamental group of a hyperbolic manifold is abelian ], 70 Thurston The Geometry and Topology of 3-Manifolds

8 4.10. INCOMPRESSIBLE SURFACES IN THE FIGURE-EIGHT KNOT COMPLEMENT Incompressible surfaces in the figure-eight knot complement. Let M 3 be a manifold and S M 3 asurfacewith S M. Assume that S S 2, IP 2,oradiskD 2 which can be pushed into M. ThenS is incompressible if every loop (simple closed curve) on S which bounds an (open) disk in M S also bounds a disk in S. Some people prefer the alternate, stronger definition that S is (strongly) incompressible if π 1 (S) injects into π 1 (M). By the loop theorem of Papakyriakopoulos, these two definitions are equivalent if S is two-sided. If S has boundary, then S is also -incompressible if every arc α in S (with (α) S) which is homotopic to M is homotopic in S to S. If M is oriented and irreducible (every two-sphere bounds a ball), then M is sufficiently large if it contains an incompressible and -incompressible surface. A 4.39 compact, oriented, irreducible, sufficiently large three-manifold is also called a Hakenmanifold. It has been hard to find examples of three-manifolds which are irreducible but can be shown not to be sufficiently large. The only previously known examples are certain Seifert fibered spaces over S 2 with three exceptional fibers. In what follows we give the first known examples of compact, irreducible three-manifolds which are not Haken-manifolds and are not Seifert fiber spaces. Note. If M is a compact oriented irreducible manifold D 3, and either M or H 1 (M) 0, then M is sufficiently large. In fact, M 0 H 1 (M) 0. Think of a non-trivial cohomology class α as dual to an embedded surface; an easy argument using the loop theorem shows that the simplest such surface dual to α is incompressible and -incompressible. Thurston The Geometry and Topology of 3-Manifolds 71

9 4. HYPERBOLIC DEHN SURGERY The concept of an incompressible surface was introduced by W. Haken (International Congress of Mathematicians, 1954), (Acta. Math. 105 (1961), Math A. 76 (1961), Math Z 80 (1962)). If one splits a Haken-manifold along an incompressible and -incompressible surface, the resulting manifold is again a Haken-manifold. One can continue this process of splitting along incompressible surfaces, eventually arriving (after a bounded number of steps) at a union of disks. Haken used this to give algorithms to determine when a knot in a Haken-manifold was trivial, and when two knots were linked. Let K be a figure-eight knot, M = S 3 N(K). M is a Haken manifold by 4.40 the above note [M is irreducible, by Alexander s theorem that every differentiable two-sphere in S 3 bounds a disk (on each side)]. Here is an enumeration of the incompressible and -incompressible surfaces in M. There are six reasonably obvious choices to start with; S 1 is a torus parallel to M, S 2 = T 2 -disk = Seifert surface for K. To construct S 2,take3circleslying above the knot, and span each one by a disk. Join the disks by a twist for each crossing at K to get a surface S 2 with boundary the longitude (0, ±1). S 2 is oriented and has Euler characteristic = 1, so it is T 2 -disk. S 3 = (Klein bottle-disk) is the unoriented surface pictured. 72 Thurston The Geometry and Topology of 3-Manifolds

10 4.10. INCOMPRESSIBLE SURFACES IN THE FIGURE-EIGHT KNOT COMPLEMENT. S 4 = (tubular neighborhood of S 3 )=T 2 2disks. S 4 =(±4, 1), (depending on the choice of orientation for the meridian). S 5 = (Klein bottle-disk) is symmetric with S 3. S 6 = (tubular neighborhood of S 5 )=T 2 2disks. S 6 =(±4, 1). It remains to show that Theorem Every incompressible and -incompressible connected surface in M is isotopic to one of S 1 through S 6. Corollary. The Dehn surgery manifold M (m,l) is irreducible, and it is a Hakenmanifold if and only if (m, l) =(0, ±1) or (±4, ±1). In particular, none of the hyperbolic manifolds obtained from M by Dehn surgery is sufficiently large. (Compare 4.7.) Thus we have an infinite family of examples of oriented, irreducible, non-haken-manifolds which are not Seifert fiber spaces. It seems likely that Dehn surgery along other knots and links would yield many more examples Proof of corollary from theorem. Think of M (m,l) as M union a solid torus, D 2 S 1, the solid torus being a thickened core curve. To see that M (m,l) is irreducible, let S be an embedded S 2 in M (m,l), transverse to the core curve α (S intersects the solid torus in meridian disks). Now isotope S to minimize its intersections with α. IfS doesn t intersect α then it bounds a ball by the irreducibility of M. If it does intersect α we may assume each component of intersection with the solid torus D 2 S 1 is of the form D 2 x. If S M is not incompressible, we may divide S into two pieces, using a disk in S M, each of which has fewer intersections with α. If S does not bound a ball, one of the pieces does not bound. If S M is -incompressible, we can make an isotopy of S to reduce the number of intersections Thurston The Geometry and Topology of 3-Manifolds 73

11 4. HYPERBOLIC DEHN SURGERY with α by 2. Eventually we simplify S so that if it does not bound a ball, S M 4.43 is incompressible and -incompressible. Since none of the surfaces S 1,...,S 6 is a submanifold of S 2, it follows from the theorem that S in fact bounds a ball. The proof that M (m,l) is not a Haken-manifold if (m, l) (0, ±1) or (±4, ±1) is similar. Suppose S is an incompressible surface in M (m,l). Arrange the intersections with D 2 S 1 as before. If S M is not incompressible, let D be a disk in M with D S M not the boundary of a disk in S M. Since S in incompressible, D = D for some disk D S which must intersect α. The surface S obtained from S by replacing D with D is incompressible. (It is in fact isotopic to S, since M is irreducible; but it is easy to see that S is incompressible without this.) S has fewer intersections with α than does S. IfS is not -incompressible, an isotopy can be made as before to reduce the number of intersections with α. Eventually we obtain an incompressible surface (which is isotopic to S) whose intersection with M is incompressible and -incompressible. S cannot be S 1 (which is not incompressible in M ( m, l)), so the corollary follows from the theorem. Proof of theorem Recall that M = S 3 N(K) is a union of two tetrahedra-without-vertices. To prove the theorem, it is convenient to use an alternate description of M at T 2 I with certain identifications on T 2 {1} (compare Jørgensen, Compact three-manifolds of constant negative curvature fibering over the circle, 4.44 Annals of Mathematics 106 (1977), 61 72, and R. Riley). One can obtain this from the description of M as the union of two tetrahedra with corners as follows. Each tetrahedron = (corners) I with certain identifications on (corners) {1}. This product structure carries over to the union of the two tetrahedra. The boundary torus has the triangulation (p. 4.11) 74 Thurston The Geometry and Topology of 3-Manifolds

12 4.10. INCOMPRESSIBLE SURFACES IN THE FIGURE-EIGHT KNOT COMPLEMENT T 2 {1} has the dual subdivision, which gives T 2 as a union of four hexagons. The diligent reader can use the gluing patters of the tetrahedra to check that the identifications on T 2 {1} are Thurston The Geometry and Topology of 3-Manifolds 75

13 4. HYPERBOLIC DEHN SURGERY where we identify the hexagons by flipping through the dotted lines. The complex N = T 2 {1}/identifications is a spine for N. It has a cell subdivision with two vertices, four edges, and two hexagons. N is embedded in M, andits complement is T 2 [0, 1). If S is a connected, incompressible surface in M, the idea is to simplify it with respect to the spine N (this approach is similar in spirit to Haken s). First isotope S 4.46 so it is transverse to each cell of N. Next isotope S so that it doesn t intersect any hexagon in a simple closed curve. Do this as follows. If S hexagon contains some loops, pick an innermost loop α. Then α bounds an open disk in M 2 S (it bounds one in the hexagon), so by incompressibility it bounds a disk in S. By the irreducibility of M we can push this disk across this hexagon to eliminate the intersection α. One continues the process to eliminate all such loop intersections. This does not change the intersection with the one-skeleton N (1). S now intersects each hexagon in a collection of arcs. The next step is to isotope S to minimize the number of intersections with N (1). Look at the preimage of S N. We can eliminate any arc which enters and leaves a hexagon in the same edge by pushing the arc across the edge. If at any time a loop intersection is created with a hexagon, eliminate it before proceeding Next we concentrate on corner connections in hexagons, that is, arcs which connect two adjacent edges of a hexagon. Construct a small ball B about each vertex, 76 Thurston The Geometry and Topology of 3-Manifolds

14 4.10. INCOMPRESSIBLE SURFACES IN THE FIGURE-EIGHT KNOT COMPLEMENT. and push S so that the corner connections are all contained in B, andsothats is transverse to B. S intersects B in a system of loops, and each component of intersection of S with B contains at least one corner connection, so it intersects N (1) at least twice. If any component of S B is not a disk, there is some innermost such component S i ; then all of its boundary components bound disks in B, hence in S. SinceS is not a sphere, one of these disks in S contains S i. Replace it by a disk in B. This can be done without increasing the number of intersections with N (1),since every loop in B bounds a disk in B meeting N (1) at most twice. Now if there are any two corner connections in B which touch, then some component of S B meets N (1) at least three times. Since this component is a disk, it can 4.48 be replaced by a disk which meets N (1) at most twice, thus simplifying S. (Therefore at most two corners can be connected at any vertex.) Assume that S now has the minimum number of intersections with N (1) in its isotopy class. Let I, II, III, and IV denote the number of intersections of S with edges I, II, III, and IV, respectively (no confusion should result from this). It remains to analyze the possibilities case by case. Suppose that none of I, II, III, and IV are zero. Then each hexagon has connections at two corners. Here are the possibilities for corner connections in hexagon A. If the corner connections are at a and b then the picture in hexagon A is of the form Thurston The Geometry and Topology of 3-Manifolds 77

15 4. HYPERBOLIC DEHN SURGERY This implies that II = I + III + II + I + IV, which is impossible since all four numbers are positive in this case. A similar argument also rules out the possibilities c-d, d-e, a-f, b-f, and c-e in hexagon, and h-i, i-j, k-l, g-l, g-k and h-j in hexagon B. The possibility a-c cannot occur since they are adjacent corners. For the same reason we can rule out a-e, b-d, d-f, g-i, i-k, h-l, and j-l. Since each hexagon has at least two corner connections, at each vertex we must have connections at two opposite corners. This means that knowing any one corner connection also tells you another corner connection. Using this one can rule out all possible corner connections for hexagon A except for a-d. If a-d occurs, then I + IV + II = I + III + II, or III = IV. By the requirement of opposite corners at the vertices, in hexagon B there are corner connections at i and l, which implies that I = II. Let x = III and y = I. The picture is then 4.49 We may reconstruct the intersection of S with a neighborhood of N, say N(N), from this picture, by gluing together x + y annuli in the indicated pattern. This yields x+y punctured tori. If an x-surface is pushed down across a vertex, it yields a y-surface, and similarly, a y-surface can be pushed down to give an x-surface. Thus, S N(N) isx + y parallel copies of a punctured torus, which we see is the fiber of a fibration of N(N) M over S 1. We will discuss later what happens outside N(N). (Nothing.) Now we pass on to the case that at least one of I, II, III, and IV are zero. The case I = 0 is representative because of the great deal of symmetry in the picture Thurston The Geometry and Topology of 3-Manifolds

16 4.10. INCOMPRESSIBLE SURFACES IN THE FIGURE-EIGHT KNOT COMPLEMENT. First consider the subcase I = 0 and none of II, III, and IV are zero. If hexagon B had only one corner connection, at h, then we would have III + IV = II + IV + III, contradicting II > 0. By the same reasoning for all the other corners, we find that hexagon B needs at least two corner connections. At most one corner connection can occur in a neighborhood of each vertex in N, since no corner connection can involve 4.51 the edge I. Thus, hexagon B must have exactly two corner connections, and hexagon A has no corner connections. By checking inequalities, we find the only possibility is corner connections at g-h. If we look at the picture in the pre-image T 2 {1} near I we see that there is a loop around I. This loop bounds a disk in S by incompressibility, and pushing the disk across the hexagons reduces the number of intersections with N (1) by at least two (you lose the four intersections drawn in the picture, and gain possibly two intersections, above the plane of the paper). Since S already has minimal intersection number with N (1) already, this subcase cannot happen. Now consider the subcase I = 0 and II = 0. In hexagon A the picture is Thurston The Geometry and Topology of 3-Manifolds 79

17 4. HYPERBOLIC DEHN SURGERY implying III = IV. The picture in hexagon B is 4.52 with y the number of corner connections at corner l and x =IV y. The three subcases to check are x and y both nonzero, x =0,andy =0. If both x and y are nonzero, there is a loop in S around edges I and II. The loop bounds a disk in S, and pushing the disk across the hexagons reduces the number of intersections by at least two, contradicting minimality. So x and y cannot both be nonzero. If I = II = 0 and x =0,thenS N(N) isy parallel copies of a punctured torus Thurston The Geometry and Topology of 3-Manifolds

18 4.10. INCOMPRESSIBLE SURFACES IN THE FIGURE-EIGHT KNOT COMPLEMENT. If I = II = 0 and y =0,thenS N(N) consists of x/2 copies of a twice punctured torus, together with one copy of a Klein bottle if x is odd. Now consider the subcase I = III = 0. If S intersects the spine N, thenii because of hexagon A and IV 0 because of hexagon B. But this means that there is a loop around edges I and III, and S can be simplified further, contradicting minimality. Thurston The Geometry and Topology of 3-Manifolds 81

19 4. HYPERBOLIC DEHN SURGERY The subcase I = IV = 0 also cannot occur because of the minimality of the number of intersections of S and N (1). Here is the picture. By symmetric reasoning, we find that only one more case can occur, that III = IV = 0, with I = II. The pictures are symmetric with preceding ones: Thurston The Geometry and Topology of 3-Manifolds

20 4.10. INCOMPRESSIBLE SURFACES IN THE FIGURE-EIGHT KNOT COMPLEMENT. To finish the proof of the theorem, it remains to understand the behavior of S in M N(N) =T 2 [0,.99]. Clearly, S (T 2 [0,.99]) must be incompressible. (Otherwise, for instance, the number of intersections of S with N (1) could be reduced.) It is not hard to deduce that either S is parallel to the boundary, or else a union of annuli. If one does not wish to assume S is two-sided, this may be accomplished by studying the intersection of S (T 2 [0,.99]) with a non-separating annulus If any annulus of S (T 2 [0,.99]) has both boundary components in T 2.99, then by studying the cases, we find that S would not be incompressible. It follows that S (T 2 [0,.99]) can be isotoped to the form (circles [0,.99]). There are five possibilities (with S connected). Careful comparisons lead to the descriptions of S 2,...,S 6 given on pages 4.40 and Thurston The Geometry and Topology of 3-Manifolds 83

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